Question

PROBLEM 1: The 45° strain rosette is mounted on the link of the backhoe. The following readings are obtained from each gauge 650[10). = -300(10) = 480(10"). Determine principal stresses on the link if it is made from steel with modulus of elasticity E - 200 GPa and Poisson's ratio v 0.3. (30p)

Answer 1

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The answer is given below.

& Strain Rosette (450) strain values Ea=650x106 Eb=-300x106 Ec= 420x106 Strain transformation equatione-> Ex= Ea Gg-Ge Eb-Chatte) Ex = 650x106 Ey= 480x106 Sxy=[2x (-300) -(650 +480) x 106 = -1730x106 Using these strains, the plane stresses are derived using Hookers Law:- е е и by = ry - USA Assuming Ty &ox are tensile stresses. E = 200GPa V=M=0.3 650x100 = (5x - 0.35y) / → 2X105x660x176-62-0.364 → 1-0.364 = 130- (1) → 480x16= (by-0.304)* – 4 oxfox 2x - 5 -0.3 - 0.36% = ->)

→ Solving equations (1)&(2) >> x = 174.505 MPa (Tension) by = 148.352MPa (Tension). Also, &xy= TaylG G E C Shear Modulus of Elasticity) 2114M) → G = 200x102 MPa 2x (1+0.3) G= 76.92 GPa → Txy= Gxkey = 76.92x10x (-1730x106) = - 133.071 MPa Principal Stresses (Major) OP = Oxtoy + 2 Jury-ox)2 + 4 tay = 295.14 MPa (Tensile) 2 (Minor) TP2 = Tatry _ I (6y-ox)² + 4 tay = 27.716 MPa